• Congratulations to the Class of 2025 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Polynomial Question! (1 Viewer)

shmickmick

New Member
Joined
Aug 27, 2005
Messages
18
Gender
Male
HSC
2006
hey just wondering if anyone could help me with a question as im the only guy doing it in my skool..............

for P(x) 2x^3 - x^2 - 6x +3

how do you find that x= 1/2 for the trial and error method i thought it could only be a factor of 3 and 2 eg 2/3 ,3/2, 1 etc

thanks........mick
 

_ShiFTy_

Member
Joined
Aug 7, 2005
Messages
185
Gender
Male
HSC
2006
You couldve have also factorised that equation

2x^3 - x^2 - 6x +3
x^2(2x - 1) - 3(2x - 1)
(2x - 1)(x^2 - 3)


shmickmick said:
how do you find that x= 1/2 for the trial and error method i thought it could only be a factor of 3 and 2 eg 2/3 ,3/2, 1 etc
k
Factor R: -1, 1, -3, 3
Factor S: 1, 2

1/2 is a possible solution...
 

shmickmick

New Member
Joined
Aug 27, 2005
Messages
18
Gender
Male
HSC
2006
ok cheers that explains it ...............pretty obvious now lol !
 

YBK

w00t! custom status!! :D
Joined
Aug 22, 2004
Messages
1,239
Location
47 |)35|< !!!
Gender
Male
HSC
2006
Shady01 said:
what if you can't factoirse this the way shifty has done? how can we find 1/2????
2x^3 - x^2 - 6x +3

that means there are several possibilities for factors

take for example (2x + 3) = 0

x = -3/2


Now you can see that the numerator (top) is a factor of the constant, and the denominator is a factor of the x's coeficient(bottom).

Therefore, we can apply this to the main problem. The numerator has to be a factor of 3, and 1 is a factor of 3 and the dinominator has to be a factor of 2, and 2 is a factor of 2.

hope that made sense :)

edit: fixed typo :D
 
Last edited:

davin

Active Member
Joined
Dec 10, 2003
Messages
1,567
Gender
Male
HSC
N/A
the basic form is that if you have some function Ax^n + (other lower powers of x) + B where A and B are just integers, all the possible factors you need to try are factors of B over factors of A, both positive and negative values.

So, for example, for:
6x^4 + 3x^3 - 5x^2 +x -8
your number for A is 6 and your number for B is -8
the factors of B are +1, -1, +2, -2, +4, -4, +8, -8
the factors of A are +1, -1, +2, -2, +3, -3, +6, -6

the possible roots, then, are all the possible combinations of B over A, in this example:
+1, -1, +1/2, -1/2, +1/3, -1/3, +1/6, -1/6, +2, -2, +2/3, -2/3, +4, -4, +4/3, -4/3, +8, -8, +8/3, -8/3
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top